Convergence of $\int^\infty_0\frac{1}{1+x^2\sin^2(5x)}\,dx$

Question

I need to find out whether the following improper integral converges:

$$\int^\infty_0\frac{1}{1+x^2\sin^2(5x)}\,dx$$

I tried two comparison tests that failed, any ideas?

Answer 1

Note that we can write

$$\begin{align} \int_0^{N\pi/5}\frac{1}{1+x^2\sin^2(5x)}\,dx&=\frac15\int_0^{N\pi}\frac{1}{1+(x/5)^2\sin^2(x)}\,dx\\\\ &=\frac15\int_0^{\pi/2}\frac{1}{1+(x/5)^2\sin^2(x)}\,dx\\\\ &+\frac15\sum_{n=1}^{N-1}\int_{(n-1/2)\pi}^{(n+1/2)\pi}\frac{1}{1+(x/5)^2\sin^2(x)}\,dx\\\\ &+\frac15\int_{(N-1/2)\pi}^{N\pi}\frac{1}{1+(x/5)^2\sin^2(x)}\,dx\tag 1 \end{align}$$

The first on the right-hand side of $(1)$ creates no issue and the third integral is easily seen to converge as $N\to \infty$.

We focus attention on the second integral on the right-hand side of $(1)$ and write

$$\begin{align} \int_{(n-1/2)\pi}^{(n+1/2)\pi}\frac{1}{1+(x/5)^2\sin^2(x)}\,dx&\ge 2\int_{0}^{\pi/2}\frac{1}{1+((n+1/2)\pi)/5)^2\sin^2(x)}\,dx\\\\ &\ge 2\int_{0}^{\pi/2}\frac{1}{1+((n+1/2)\pi)/5)^2 x^2}\,dx\\\\ &=\frac{20}{(2n+1)\pi}\arctan((n+1/2)\pi^2/10)\\\\ &\ge \frac{20}{(2n+1)\pi}\left(\frac{\pi}{2}-\frac{1}{(n+1/2)\pi^2/5}\right) \end{align}$$

Inasmuch as the harmonic series diverges, the integral of interest diverges likewise.